Question

evaluate.\
\\( \int \left(13x^{4/3} - 8x^{6/5}\
ight) dx \\)\
\\( \int \left(13x^{4/3} - 8x^{6/5}\
ight) dx = \square \\) (type an exact answer.)

Explanation:

Step1: Integrate term by term

Use the power rule for integration: $\int x^n dx = \frac{x^{n + 1}}{n + 1} + C$ (where $C$ is the constant of integration, but since we are asked for an antiderivative, we can include it or not; here we'll include it at the end). First, integrate $13x^{4/3}$:
The coefficient is 13, and $n = 4/3$. So, $\int 13x^{4/3} dx = 13 \times \frac{x^{(4/3) + 1}}{(4/3) + 1}$.
Calculate the exponent: $(4/3) + 1 = 7/3$. So this becomes $13 \times \frac{x^{7/3}}{7/3} = 13 \times \frac{3}{7}x^{7/3} = \frac{39}{7}x^{7/3}$.

Step2: Integrate the second term

Now integrate $-8x^{6/5}$. The coefficient is -8, and $n = 6/5$. So, $\int -8x^{6/5} dx = -8 \times \frac{x^{(6/5) + 1}}{(6/5) + 1}$.
Calculate the exponent: $(6/5) + 1 = 11/5$. So this becomes $-8 \times \frac{x^{11/5}}{11/5} = -8 \times \frac{5}{11}x^{11/5} = -\frac{40}{11}x^{11/5}$.

Step3: Combine the results and add the constant

Combine the two integrated terms and add the constant of integration $C$:
$\int (13x^{4/3} - 8x^{6/5}) dx = \frac{39}{7}x^{7/3} - \frac{40}{11}x^{11/5} + C$

Answer:

$\frac{39}{7}x^{\frac{7}{3}} - \frac{40}{11}x^{\frac{11}{5}} + C$ (If the problem allows omitting the constant, it can be $\frac{39}{7}x^{\frac{7}{3}} - \frac{40}{11}x^{\frac{11}{5}}$)